leetcode: 刪除鏈表中的冗餘結點

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
Example 1:
Input: 1->2->3->3->4->4->5
Output: 1->2->5
Example 2:
Input: 1->1->1->2->3
Output: 2->3

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
#include<unordered_map>
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head == nullptr)
            return nullptr;
        
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        
        ListNode* prev = dummy;
        ListNode* curr = head;
        
        
        while(curr != nullptr)
        {
            while(curr->next != nullptr && prev->next->val == curr->next->val)
            {
                curr = curr->next;
            }
            
            if(prev->next == curr)
                prev = curr;
            else
                prev->next = curr->next;
            
            curr = curr->next;
        }
      
        return  dummy->next;
    }
};